advanced-algebra

[rose mirage]

I think that's FAR easier than this exercise lol

[robust rain]

okay, I think I'll move on for now, and try to decipher this when I know what a natural transformation is lol

thank you wew for the guidance

[rose mirage]

if you want to come back to this exercise before reading on, I'd try and prove first that tensor and hom are adjoint and then try and prove this exercise for those functors specifically

cause the technique should be the same

[robust rain]

gotcha. I'll see how much pain/bookkeeping hom-tensor adjunction causes me

[hallow bone]

should be fine

it's just slightly tedious

[fierce steeple]

Sorry i meant like were you able to do it

For 2.3.1 I think this is just commuting things with colimits/limits

[subtle smelt]

Not quite, I am doing to looking at
[\begin{tikzcd}[cramped,sep=scriptsize]
{\underset{\times f}{\varprojlim} M} && M \
\
0 && {\underset{n}{\varprojlim} \operatorname{cofib}(M \xrightarrow{f^n}M)}
\arrow[from=1-1, to=1-3]
\arrow[from=1-1, to=3-1]
\arrow[from=1-3, to=3-3]
\arrow[from=3-1, to=3-3]
\end{tikzcd}]

[broken turtle]

Kerr

[fierce steeple]

I guess the details depend on how Adeel defined the quotient by an element of pi_0(A). I would guess it's defined as the cofibre of multiplication by that arrow

[subtle smelt]

might just be routine shift fiber/cofiber stuff from here, but I need to get used to the details

[fierce steeple]

I think here you can view this as a limit of (co)fibre sequences

[subtle smelt]

Yeah I am going back to understand what actually the diagram of the M - f^n - > M is to proceed from there

M/f^n is defined as M \otimes_A (A/f^n), where the latter is defined as A \otimes_Z[X] Z where f^n in pi_0(A) is identified with Z[X) -> A.

[fierce steeple]

Sure ye

[subtle smelt]

Okay the result looks tautological, which is good. lim_n M -> M -> cofib(lim_n M -> M)

Can I throw a short cocart fibration question at you potato 👉 👈

[lone jacinth]

So it's essentially the yoneda lemma.

Notice that for any g: F(A) -> B you can make a naturally square
Mor(FA, FA) -> Mor(FA, B)
V V
Mor(A, GFA) -> Mor(A, GB)

Now just fill in what the maps are in this square and follow the identity in FA

[robust rain]

Can I be honest with you

After I asked about the yoneda lemma, I thought about it for about 30minutes more, decided I couldn't be bothered and moved on.

I promise that is unusual

I will go back to this soon

[robust rain]

It brings me great shame to admit this

[sinful turtle]

,, Phi_E = k_e q r int_{-b/2}^{b/2} int_{-l/2}^{l/2} frac{dx dy}{(x^2+y^2+r^2)^{3/2}}

[broken turtle]

Weinnion

[robust rain]

In order for this pair to be adjoint, the functor $(\cdot) \otimes_B A$ has to go from $\text{Mod}_B \to \text{Mod}_A$. Maybe I'm being stupid, but how are we putting an $A$-module structure on $N \otimes_B A$, where $N$ is a $B$-module?

okay I am being stupid

It goes from Mod_B to Mod_A

[subtle smelt]

B -> A being your ring homorphism is disturbing

[robust rain]

I can't read

Wait no I'm not

The functor goes from Mod_B to Mod_A, which means we need an A-module structure on N \otimes_B A

[broken turtle]

swifteeee

[hallow bone]

yes

[digital parcel]

a(n otimes a’) = n otimes aa’

[hallow bone]

usually it's written A \otimes_B N, though

[digital parcel]

This is just extension of scalars

[hallow bone]

which makes the module structure more obvious

[robust rain]

Oh I was thinking about how to make a multiplication n *a

I'm so silly

I've done this before

Thanks guys

[hallow bone]

you don't, that's what the tensor product is for :>

[robust rain]

xd

[hallow bone]

it's like, oh you don't have an A-module structure? We'll GIVE YOU ONE

and sometimes the module dislikes it so much that it collapses to triviality

[robust rain]

ya ya ya

I usually know this I think, just have spent too much time thinking today

[hallow bone]

that's felt

maybe take a break lol

that is also okay sometimes

[robust rain]

yes soon

[subtle smelt]

A common principle of most of the early🐈-stuff is about "find the most obvious thing that works" and usually nothing else could work without it having a counterexample in some heinous contrived category

[robust rain]

Yes haha. I've found that considering the identity map is usually a good way to construct other maps as the identity is the only thing you know exists for sure

[subtle smelt]

For the tensor product, you might as well pretend that your base module is xenophobic and only lets B act on it. Luckily you can get an action on A \otimes_B M regardless

[hushed bone]

Why are you letting the B algebra be called A

This is deeply troubling

[robust rain]

I'm not comfortable enough with this material to anthropomorphise my modules, but I'll keep this remark in mind for when that day comes

[subtle smelt]

Apparently that was Vakil

His soul was lost to the temptation of writing Spec A -> Spec B is my guess

[robust rain]

I've been working for about two hours and I've read one page. Axler would be so proud of me.

[subtle smelt]

You can rigidify the principle. Let B be an arbitrary and A an arbitrary B-algebra (🤢), what's your favourite B-module?

(a really simple B-module that always exists)

[hushed bone]

0

[robust rain]

the two I can think of are just the ring B, and we can equip A with a B-module structure via the map B -> A.

which is what is meant by A being a B-algebra ofc

[hushed bone]

You forgot 0

[robust rain]

and the 0 module

thank you

[hushed bone]

[subtle smelt]

That's the setup of the restriction of scalars functor I think

For extension of scalars we want an A-module structure on A \otimes_B B

[hallow bone]

https://mathoverflow.net/a/2589

in this answer, theyre talking about the quantum double of a group being involved in the representation theory? also the stuff about double loop spaces, and sheaves on that having a certain E[2] structure, whatever that mean

what are some resources about these things? seems interesting

[fierce steeple]

Loop spaces are a topic in alg topology / htpy theory and there's lots of stuff about them but not sure what ur background is with such things

This fits into a big story and like these E_n operads are very important

[hallow bone]

thh not enough probably, but this seems like good motivation to learn

like quantum doubles and stuff are things im super interested in

[hallow bone]

right ive heard of those

E_\infty rings are important or smt

[fierce steeple]

Pog

[fierce steeple]

Ye this is one of the main generalisations of commutative rings in htpy theory and stuff

[hallow bone]

i see

i guess i just need to start learning homotopy theory then?

maybe with a focus towards representation theory as seems to be alluded to by that answer

no clue how or where to start though

[sinful turtle]

Y=(X-1)(X-2)(X-3), find X for which Y ≥ 0

HARDEST ALGEBRA

Harvard University Question

[shy cargo]

And you get an —

[robust rain]

Quick clarification of Ad1. When we say that Hom(A,B) is an abelian group that commutes with composition we mean: 1. we can define an abelian group structure on Hom(A,B), and 2. composition of functions distributes over + h(f+g) = hf +hg. Is this the right interpretation?

[last talon]

h(f + g) = hf + hg and (f + g)h = fh + gh

(Ie, it distributes in both directions)

And it’s not “we can define an abelian group structure on”
It’s “we have an abelian group structure on”

(Ie, the data of the abelian group structure on Mor(A, B) is part of the data of an additive category )

[robust rain]

Okay I see thank you

What does vakil mean by 0_Z here? (the other two axioms in the definition of additive category are that a 0 object exists, and products exist for any pair of objects in C.

[hard kite]

Let $R \subset S$ be an extension of domains and $\mf{a}$ an $R$-ideal. Throughout my life I have always assumed that $S \otimes_R \mf{a} = S\mf{a}$, but I am beginning to doubt this. Certainly this is true whenever the extension is flat. Is this always true, and, if not, are there weaker conditions that are sufficient?

I am dealing with the following situation: $R$ is Dedekind with fraction field $K$ and $S$ is the integral closure of $R$ in $L$. Then $S$ is Dedekind and has finite fibres and residual extension degrees over $R$. I believe this is enough to conclude that $S$ is flat over $R$.

This is what I came up with: Let $\mf{p}$ be a prime of $R$. It suffices that $S_\mf{p}$ is flat over $R_{\mf{p}}$, thus suffices that projective, suffices that it is finite since $R_\mf{p}$ is a DVR. By Nakayama it suffices that $k_\mf{p} \otimes_{R} S_\mf{p} = \prod_{\mf{P} \supset \mf{p}} S/\mf{P}^{e(\mf{P}/\mf{p})}$ has finite $k_{\mf{p}}$-dimension. We have a finite filtration [
0 \subset \mf{P}^{e-1}/\mf{P}^e \subset \dots \subset S/\mf{P}^e
]
and the quotients are all isomorphic to $k_{\mf{P}}$, which is finite over $k_{\mf{p}}$. Hence, $k_\mf{p} \otimes_{R} S_\mf{p}$ is finite over $k_{\mf{p}}$.

This all seems a bit contrived for such a simple situation, are there better ways to see it?

[last talon]

The morphism Z -> 0 -> Z

[broken turtle]

1728

[robust rain]

ah thank you. and if Z is the 0-object, then id_Z and 0_Z give two morphisms Z to Z, hence must be equal as there is a unique morphism Z -> Z

[last talon]

Yes

[robust rain]

thanks!

[hard kite]

just some extra context, this came up when trying to understand the isomorphism between the ideal class group and the picard group of a Dedekind domain. Specifically, I want to understand why the induced maps coincide. For this reason would expect this to be true more generally, not just when S is the integral closure in some finite extension.

[lone jacinth]

It's equivalent to
Tor_1(S, R/a) = 0

[hard kite]

oh right

I am wondering, do you know concrete conditions besides flatness of S that give this? In the case when working with an (integral?) extension of Dedekind domains, is this always satisfied? Do you think the proof i outlined is overkill?

[lone jacinth]

So if you want it to hold for all a it should be iff S is flat.

[hard kite]

right this is because S is flat over R iff every inlcusion of ideals in to R remains injective after tensoring

hmmm maybe morphisms of Dedekind domains are always flat then or something

I knwo that S does nto need to be finite over R i will look into that

[drowsy niche]

injective ones are

[hard kite]

why?

[drowsy niche]

bc every torsion-free module over a dedekind domain is flat

[hard kite]

every finite torsion-free module over a Dedekind domain is flat. But I believe that S does not need to be finite

[drowsy niche]

iirc it holds in general (?)

[hushed bone]

No, all torsion free modules are flat

[drowsy niche]

well I know it holds for PIDs and flatness is local

[hard kite]

S not being finite is so cursed btw lol

[hushed bone]

Besides a torsion free module is a direct limit of finite torsion free modules

[hard kite]

I always came across another cursed extension recently: there exists a quadratic extension L/K of number fields such that O_L is not monogenic over O_K

[astral ginkgo]

Does Tor(S,R/a)=0 for all a imply S is flat?

[past cove]

Yes

You can even restrict to f.g. a

[astral ginkgo]

a here is an element

[past cove]

Ohhhh

[astral ginkgo]

I know the statement for (f.g.) ideals, I'm wondering if it's enough to work with principal ideals

In generally it surely can't be(?), but maybe over dedekind domains you somehow can

[lone jacinth]

a was an ideal above, but if you're asking for a an element I would assume no

[astral ginkgo]

Oh, my bad. I do think it's true in this context though, because flatness is a local property and local dedekind domains are PIDs, and tor commutes with localisation

[hushed bone]

No, if you can do this then every torsion free module would be flat

By looking at 0 -> A -a> A -> A/a -> 0

[astral ginkgo]

Ok yea, which is why it works for dedekind domains since every tf module is flat there

[hushed bone]

And then tensoring with a torsion free module M

Sorry I wrote the wrong sequence

You use the usual 0 -> (a) -> A -> A/a -> 0

[astral ginkgo]

You want (a)->A

Yea

[hushed bone]

And you use the fact that since (a) is principal that M (x)_ A has all its elements of the form m (x) xa, simple tensors

This proof is how you show any ring where you have fg ideals are principal have flat iff torsion free

This includes valuation rings

[hard kite]

wait i cant apply Nakayama exactly because I dont know that S_p is fg lol

is this even true then ?

hmm

[hushed bone]

What is L first of all?

[hard kite]

it is a quadratic extension of K

[hushed bone]

I don’t know if quadratic forces S to be finite over R but in general it isn’t for a general finite extension L

[hard kite]

it does not

[hard kite]

.

[hushed bone]

Oh

What are you asking is true or not?

[hard kite]

wether S is flat as an R-module

[hushed bone]

dude we’ve been saying it is for ages

It’s a torsion free module so it’s flat

[hard kite]

but its not finite

[hushed bone]

Everyone has been telling you that doesn’t matter lol

[hard kite]

really : [

huh

[hushed bone]

It’s a direct limit of its finitely generated submodules which are torsion free

For one

[hushed bone]

Or you can go down to the localizations and then run this proof

[hard kite]

oh sorry i guys i kind of tapped out when i saw direct limit my bad

i will look into thank you chat !

[hushed bone]

I mean I told you it’s true without finite ages ago without even saying that lol

[hushed bone]

.

[hard kite]

ooooo

i think i halucinated the word not in that sentence

Thank you, this is really helpful. And sorry for misreading it and wasting your time.

[drowsy niche]

let k be a field, and F a free group considered as an ordered group under some < (e.g. using the Magnus embedding proof or some other method).

Let D = k((F,<)) be the relevant Mal'cev-Neumann ring. If M is any left k[F]-submodule of D, is there some way of proving that Tor_1^kF = 0?

a few relevant things here: k[F] has global dimension at most 1

so it would suffice to show that Tor_1^kF = 0 by looking at the Tor LES for
0 -> M -> D -> D/M -> 0

[vestal moss]

wasnt there someone here who talked about quantales at some point

i just learned about them recently when i was reading a paper on effect systems for programming languages

[near lantern]

BTW I think I worked out at least for abelian groups that this is false (but only because the statement has to be fixed).

To be more precise, let G be a finite group and R a Dedekind domain with |G| invertible such that all irreps of G are defined over K = Frac(R) and in fact defined over R (without using the Hilbert class field (or whatever analogue is available in general) to make the lattice principal). For example, if G is finite and K contains all exponent(G)^th roots of unity, then so does R (because those roots of unity are integral over ℤ hence over R), so all irreps of G over say ℂ are defined over R. Then the isomorphism of G with a product of matrix algebras (not over division algebras) is defined and an isomorphism over R. In particular, G-reps are the same as k := (#irreps) different R-modules, and going from those to the same number of K-vector spaces kills something like Pic(R)^k inside the Grothendieck group.

[eager hound]

The problem was going from M(L) to M_n(O_K) (where L is a lattice in O_K^n) still not really sure how you’re getting past this step in your argument

here M(L) is just End(L)

some order in M_n

[near lantern]

I didn't. Just realised the statement I guessed was somewhat off even when we have O_K^n.

[eager hound]

ah I see, yes taking K_0 doesn’t necessarily help kill the class group

[near lantern]

Does K0 have some kind of Pic(R)-module structure where e.g. I acting on the trivial rep R^n gives R^{n-1} (+) I?

You could do (- (⨯) I) but applied to R^n that gives I^n = R^{n-1} (+) (n[I]).

I think if the right such action exists then we can say: K0 over both R and K splits into blocks corresponding to irreps of G (over K) since the central idempotents are defined over R. On each block the map of K0_R to K0_K should be taking coinvariants for this action.

[near lantern]

IG this is the only additive (wrt (+) on K0) way to act unfortunately.

[eager hound]

Why not just write it as an exact sequence?

Oh I see it’s probably not

[near lantern]

I guess what I am asking is essentially: can I do that and describe the kernel in an "intrinsic" way?

[eager hound]

Yeah I thought it would just be span(Pic(R)) but I guess it’s possible that it’s not

[near lantern]

As I said above, if G has k irreps over K the kernel should look like approximately Pic(R)^k, not Pic(R).

[eager hound]

why?

[near lantern]

Suppose G = prod of k matrix algebras is valid over R. Then K0(RG) = K0(R)^k, K0(KG) = K0(K)^k = ℤ^k and the map is K0(R) → K0(K) in each component. It is surjective (as you already proved in general by the lattice argument), but its kernel is Pic(R)^k.

[eager hound]

True okay but this does give an explicit description of the kernel no?

So what’s unsatisfactory about that

[near lantern]

Yes, but I feel it would be nicer for it to be more "intrinsic". Not too sure what I mean, but e.g. I think this description isn't manifestly invariant of the choice of lattice.

1 min let me generalise a bit first. I think we can work with any lattice.

Let V be an irrep over K and L a G-stable lattice. Restrict to reps with only V-isotypic component (this can be done over both R and K using central idempotents or other methods). Then (assuming for simplicity that no division algebras show up, i.e. End_G(V) = K) these categories are M(L)-mod and M(V)-mod, which are Morita-equivalent to R and K respectively, via L' ↦ L (⨯) L' and V' ↦ V (⨯) V'. That is, any G-rep over R with only V-isotypic component is L (⨯)_R L' for a unique R-module L', and similarly for K and V. In terms of these equivalences, M(L)-mod → M(V)-mod becomes the usual extension of scalars R-mod → K-mod. In particular the kernel is still Pic(R) (for each V, so Pic(R)^{#irreps} overall).

[eager hound]

I mean isn’t the point just that any lattices will induce a morita equivalence from K_0(R)^k to K_0(\prod_i End(L_i))

[near lantern]

Let's say by "intrinsic" I want to see a G-rep M over R and tell whether it's in the kernel "without choosing lattices", maybe just by taking isotypic components and checking what the isotypic components are in K0(R).

Or at least I have an intuition that this ought to be possible and if it's not, I want to understand why.

[eager hound]

Didn’t we answer this question explicitly in the above description?

[near lantern]

Ah wait actually is it true that RG is isomorphic to ∏_i End(L_i)? Over a field one proves this using the Jacobson density theorem, using that L_i is actually a simple module of the ring. Here it's just indecomposable projective.

[eager hound]

Since we are assuming that RG contains all of the central idempotents yes

[near lantern]

I think maybe not? We need to take an isotypic component M_V, write it as L (⨯) L' where L is a lattice in V, and check L' in K0(R). Finding L' seems not-so-explicit.

Hmm but L' = Hom_RG(L, M_V) probably.

[eager hound]

A class [P] - [Q] will be in the kernel iff P_K,Q_K have the same multiplicities of every irrep

[near lantern]

Right, so it's a product of factors acting on each Li. But why does that factor map onto End(Li)?

[eager hound]

so the explicit description is just via the character

[eager hound]

Well because the endomorphism algebras of lattices are exactly the maximal orders in M_n(k)

And since our original algebra was maximal every projection onto each factor must be maximal (otherwise it’s trivial to make a new order which contains it)

[near lantern]

This is by the same make-a-stable-lattice argument?

[eager hound]

Yeah

[near lantern]

Why is RG maximal?

[eager hound]

If you mean the implication maximal implies endomorphism of lattice

[eager hound]

one can prove this by hand? But is that relevant to what we’re talking about?

[near lantern]

You used it to prove the factors of RG are maximal and therefore equal to End(Li).

[eager hound]

Okay well this boils down to the same question for dvrs where is is not so hard to see

[near lantern]

OK, I understood something else that was confusing me. An M_n(R)-module which is rank n as an R-module cannot be of the form R^{n-1} (+) I (unless I is n-divisible in Pic(R)). It has to be of the form I^n = R^{n-1} (+) (n ⋅ I in Pic(R)). So the kernel of M_n(R)-mod → M_n(K)-mod is really generated by [R^n (⨯) I] - [R^n], this doesn't miss anything like [R^{n-1} (+) I] - [R^n].

[flat zephyr]

The channels said algebra 💔 what am I looking at

[near lantern]

OK, having seen that you can just repeat the same things as for fields with (i) the make-a-lattice argument (ii) the Morita equivalence argument, the essential content of 1/|G| ∈ R making the rep theory the same as over K probably lies in the fact that RG is a maximal order. E.g. ℤC_2 is not a maximal order since it is contained in {a+bg: a, b ∈ 1/2 ℤ, a+b ∈ ℤ}.

[eager hound]

Actually there’s a very nice proof, suppose X is an order in KG such that RG is contained in X, then take the smallest m such that \pi^m is contained in RG. Let Y be this multiple of X. Then Y \cap \pi RG \subset Y \subset RG. Note that Y is a two-sided ideal of RG. Now if we reduce modulo \pi we find that Y mod \pi is a square zero ideal of RG/\pi which is semisimple. So Y mod \pi is a square zero ideal, so it is trivial so X does not exist because this contradicts the minimality of m

[near lantern]

Let me try for a bit to work this out myself.

[eager hound]

this proves maximality when R is a dvr so one gets it when R is global

by localizing

[near lantern]

OK if I give up I'll read this

[eager hound]

Yeah that is the difference one gets by inverting this idempotent, otherwise as you say RG will be some ugly gluing of orders on different summands in general

[near lantern]

Hmm if we replace ℤC_2 by this maximal order we should get back all the nice properties though, right?

So the difference between rep theory of G over O_K and K should be the same as the difference between (projective module theory for) the order O_K G and some maximal order it's contained in.

[near lantern]

This is algebra 🥺 ❤️‍🩹

Does it get more algebra than talking about modules over a ring?

[eager hound]

the telos of algebra is studying relationships between certain stable infinity categories

actual algebras in sets are simply an object-level distraction

[near lantern]

IG I haven't done any algebra in my life yet 😔

Got to have patience

[hushed bone]

Talos save us

[astral maple]

My disappointment is immeasurable and my evening is ruined.

[eager hound]

Not to be nG or anything but it’s a reference to the introduction of this https://people.mpim-bonn.mpg.de/scholze/Exp1252_Scholze.pdf

[narrow kraken]

i can't figure out to how to show this, i tried just brute-forcing this but it became messy really fast. i also tried using the fact in the image below, but the problem is R does not contain the ring of scalar k[x]-matrices.

[eager hound]

The sub ring of diagonal matrices inside of R generates a ring isomorphic to k[x] though

The original R I mean

[astral ginkgo]

But a_11 must be in k

[eager hound]

yes

but none the less what I said is true

[astral ginkgo]

Oh diagonal

Oops I thought you said scalar

[eager hound]

Oh I see

Yeah I just mean a_{12}=0

[astral ginkgo]

Yea my thought was also just that it contains an isomorphic copy of k[x]

[eager hound]

Then the whole ring is a fg right module over these matrices

[lone jacinth]

So the easy way is just to notice that every ideal is in particular a k[x]-submodule, and as R is a Noetherian k[x] module, all the ideals are fg.

As for the explicit construction, an ideal is either
[0, f]
[0, xg]
where (f, g) is in a k[x]-submodule of k[x]^2
or
[g(0), f]
[0, g]
where g is in some ideal in k[x] (not contained in (x)) and f is arbitrary.

You see this because if g(0) is nonzero, then
[g(0), f] [0, h]
[0, g] [0, 0]

[0, g(0)h]
[0, 0]
so if the ideal containes such an element with non-zero g(0) you can get any f

And if g(0) is always 0, it just reduces to a k[x] module

[last talon]

What are some good examples of non-commutative rings?
So I know of:

1. Group Rings (the only real example 💜 )
2. Weyl Algebras
3. Non-commutative polynomial rings
4. Endomorphism rings (Inc. matrix algebras)
5. Universal Enveloping Algebras

[astral maple]

Quaternion algebras and their orders.

[last talon]

What are those?

[hidden shore]

graded-commutative rings

[last talon]

True
they’re basically commutative though…

[astral maple]

The “classical” quaternions are generated over R by two elements i, j with the relations i^2=j^2=-1, ij=-ji.

You can extend this construction over a general base field K (say of characteristic not 2) by picking two elements a, b of K and considering the K-algebra generated by two elements i, j with i^2=a, j^2=b, ij=-ji.

When K is a local or global field, the arithmetic of this algebra is very rich!

[hidden shore]

https://media.discordapp.net/attachments/1000143691421331626/1496238282365730856/togif-10.gif

[weak lodge]

How about Lie rings?

[hallow bone]

what are those?

[weak lodge]

Lie algebras over Z

[last talon]

Are those the same thing as Universal Enveloping Algebras?

[hallow bone]

quantum deformations of universal enveloping algebras

[hallow bone]

not associative

[weak lodge]

Ah were we only looking at associative rings? Sorry

[hallow bone]

quantum deformations of the matrix rings for that matter

[hallow bone]

thats not right, i meant quantum deformations of the multiplicative monoid of matrices and the subgroups GL and SL seen as an algebraic variety

[near lantern]

Division rings. Path algebras of quivers. Convolution algebras of posets. Diagram algebras like the Temperley-Lieb algebra. (For that matter, category algebras, which include the last three as well as group and matrix algebras.) Hecke algebras. Deformations of most of these.

[hallow bone]

how does one deform a general (noncommutative) algebra

genuine question

[fierce steeple]

Depends what you mean by deform lol

[wheat meadow]

what the fuck are these meant to be \genq

[wheat meadow]

crossed products

[hallow bone]

peakest

[astral ginkgo]

A general construction is the Crossed Product

FUCK

Sniped

[wheat meadow]

haha get scooped

[near lantern]

Find a basis over say a commutative algebra A, change scalars to A[t], change structure coefficients to polynomials which specialise to the original when t = 0 and are associative.

[hallow bone]

not A[[t]]?

[near lantern]

You can classify these by the right cohomology or something, giving you a space of possible deformations.

[last talon]

I did a course on this last year and everything’s coming back 🙃

[wheat meadow]

also me when my thesis topic \silly

well i am interested in more than just orders in quaternion algebras but they show up!

[fierce steeple]

Nice

[hallow bone]

omgah

[near lantern]

Oh, that too, I guess. Perhaps any commutative ring B with an ideal I in its Jacobson radical such that B = A (+) I.

[fierce steeple]

More generally if ur given a map A -> B of commutative rings and an associative B-algebra u can ask if there's an A-algebra which base changes to what u started with

[hallow bone]

oh descent?

[last talon]

If there was a non-commutative algebra person in Cambridge who offered summer research in either of the last two years, there’s a good chance I’d be a non-commutative algebra person

[near lantern]

Isn't that asking when your deformation is constant?

[fierce steeple]

Ig that is one way to state it but I would just call this a deformation or lift of ur ting

[fierce steeple]

I'm not sure what you mean

[near lantern]

Not the best illustration of deformation, I would think.

I mean if the B-algebra is the base-change of an A-algebra you are not deforming the A-algebra.

[fierce steeple]

Well this is what it means to deform an associative algebra at least in one sense (the sense that is controlled by Hochschild cohomology)

[fierce steeple]

You are deforming the B-algebra

[hallow bone]

ive lowkey been looking for stuff on deformations of noncommutative algebras but idk i cant really find stuff i guess i dont know where to look

[near lantern]

Oh lol

[fierce steeple]

You example is taking like B[[t]] -> B or B[t] -> B

[near lantern]

Ah of course one wouldn't want to assume a section so one can do mixed-characteristic

mb

[fierce steeple]

Ig my bad cause my use of A conflicted with yours aha

[last talon]

https://arxiv.org/pdf/1212.0914

[fierce steeple]

Also like ig usually this is not part of the data that is actually needed

[wheat meadow]

yoo travis mention

[astral ginkgo]

Given a Finite Galois Extension K/F with group G, You can first form the twisted group ring K_t[G], defined by

(xg)(yh)=(xg.y)(gh)

Where g acts as the galois action, g,h in G and x,y in K.

Now, this is always a K- Central Simple Algebra (unlike regular Group Rings). You can further twist this via cohomology: Given a 2-cocycle f\in H^2(G,K^x) you can twist multiplication further via

(xg)(yh)= (xg.y f(g,h))(gh)

This is ALSO a central simple algebra, and up to Brauer Equivalence, these are all the Central Simple Algebras, but not up to isomorphism.

An important question is finding noncrossed products (in particular noncrossed product division algebras)

[wheat meadow]

wait mico do u know ariff

[hallow bone]

noncomm geometry
scary..
but i will be brave!!

[last talon]

Nope

[fierce steeple]

Was this at your uni? Cool they offer a course on this

This is stuff I worked on last year lol

[wheat meadow]

cam doing cam things lol

[last talon]

Yup
There’s a guy here who offers some insane topics courses in non-commutative and/or group-y stuff each year

[fierce steeple]

Except I wonder how much was done not in char 0

[last talon]

(He did topics in infinite groups this year)

[fierce steeple]

Nice

[wheat meadow]

ah fair, he's doing part iii this year but he's coming back to imperial to do phd with travis iirc

[astral ginkgo]

Everyone please take a moment to appreciate crossed products, and also noncrossed products

[wheat meadow]

i have taken a great deal of moments to appreciate crossed products

so many of them in fact

[fierce steeple]

Ah was this brookes

[last talon]

Yup

[astral ginkgo]

You're so based

[wheat meadow]

i know right :3

[hallow bone]

im coming up on that chapter in kassel rn

[wheat meadow]

it's kinda sad that for all the interesting noncommutative algebra i read for my thesis, i will almost exclusively actually talk about matrix rings

but this is where the examples are so

[last talon]

It covered all characteristics where it isn’t much different from char 0, but was definitely not against taking char 0 assumptions if it made things easier

[hallow bone]

well certainly every fd algebra is a subring of a matrix ring

[fierce steeple]

Yeah

[hallow bone]

thats something

[fierce steeple]

I mean the Hodge decomposition just doesn't happen otherwise

Well in the strongest form anyway

[wheat meadow]

and all the algebras we understand are actually matrix algebras over division rings anyway because artin-wedderburn lol \silly

[hallow bone]

ah of course

every module is projective

[wheat meadow]

no but every module we understand is projective

:3

[hallow bone]

what I don't understand doesnt exist

[wheat meadow]

me when my category doesn't have enough injectives because injective modules are scary and weird and fucked up so i threw them out

[astral ginkgo]

I actually don't know Kassel

[hallow bone]

its cool, very knot theory

[wheat meadow]

knot theory scares me

it's either elementary playing with diagrams or the craziest and most cursed shit imaginable

[last talon]

Knot theory also scares me

[hallow bone]

well luckily you can just accept Reidemeister's theorem and go from there

[ornate atlas]

There was a knot theory course here this year and it looked very very boring and dry

[hallow bone]

(which is what I do)

[near lantern]

Wait, the twisted group ring is just End_F(K) = M_{|G|}(F), right?

[last talon]

I’ll probably wanna learn some eventually but the problem is that I struggle to get past the boring combi-ish parts at the start of every knot theory book/course

[ornate atlas]

I do want to learn some knot theory but it was in my already very busy semester and looked kinda shit

[wheat meadow]

this happens to a lot of cool math subjects and i really hate it -- a lot of those courses would be infinitely better if lecturers didn't feel the need to build all the technical theory and focused instead on working with interesting examples

[ornate atlas]

I honestly think this a lot and it makes me feel like a bit of a fraud at times lol. Sometimes I think courses should worry a little less about some proofs and spend more time getting you to actually do stuff

[near lantern]

Ah, probably this is supposed to be a group isomorphism H^2(G, K^⨯) → Br(K | F), so that isn't an accident.

[astral ginkgo]

Yes

[wheat meadow]

also like... a lot of the time it's just a waste of time -- i do not need to prove the isomorphism theorems for a fifth time in my lie algebras course

nobody in the room needed that

[astral ginkgo]

It is le.trivial cocycle

Aka, the identity element

[ornate atlas]

Like I think there’s at least 4 or 5 lectures of my cohomology class this sem which were just really technical proofs which sure of course are important to have done so our theorems actually work but like, I don’t think they’re very enlightening or interesting and there’s really no point to waste any time doing them in lecture

[near lantern]

I hope I learn CSAs soon 😔

[ornate atlas]

Well yeah this is true lol, the first 5 weeks of my Lie algebras course could’ve been a single lecture

[fierce steeple]

Step 1 is use another acronym

[near lantern]

Still haven't read why they are classified by cohomology

[last talon]

Yeah our lecturer didn’t do it
Although I’d’ve appreciated it if he said “look, you should be able to figure out what an ideal/subalgebra/etc is, and that the iso theorems hold, so I’m not gonna do them” instead of just assuming we’d realise he wanted us to do that

[astral ginkgo]

There's 2 ways to do this also!

The crossed product construction I described classifies them via H^2. Galois descent helps classify them via H^1

[near lantern]

Hmmmm

[astral ginkgo]

You can also prove this via étale cohomology if you're so inclined

[wheat meadow]

this is one thing i really appreciate about my elliptic curves course lol, a lot of boring technical stuff was smoothly elided in favor of the interesting and insightful

[near lantern]

Galois descent should give H^1(Aut(M_n)) = H^1(PGL_n)

[wheat meadow]

also the homeworks made us do examples but in interesting ways

[ornate atlas]

Let’s move to #1203471755449073774 maybe to let shin and raghuram talk

[near lantern]

IG the distinction is whether you fix the dimension or no

[fierce steeple]

Ah is this the like boundary map between PSL and Gm thing

[wheat meadow]

that it does (though then you can 0 -> k* -> GL_n -> PGL_n -> 0, and cohomology and theorem 90 gives you H^2(k*) it's really cute)

[fierce steeple]

I LOVE THEOREM 90

[near lantern]

I tried that but there's a H^2(GLn) in the way

[wheat meadow]

thm 90 go brr

[subtle plaza]

Vanishes by Gilbert theorem 90

[astral ginkgo]

Yes

Then yoy take a direct limit over n

Then a direct limit over finite galois extensiond

[near lantern]

NGL I still find it impossible to think of Theorem 90 any other way than that the twisted group ring is the matrix ring.

[astral ginkgo]

This naturally introduces the brauer equivalence

[astral ginkgo]

Yes that is how I think about it, and separately there's hilbret's lame theorem for nerds about norm 1 elements

[fierce steeple]

There's the perspective that coho of Gm doesn't care whether u use étale or Zariski

This is how I think of it

[near lantern]

That's true for any scheme?

[fierce steeple]

Yes

[near lantern]

And H^m(GLn) or just H^1(GL1)?

[fierce steeple]

Though I stated it informally lol

[astral ginkgo]

Btw you can prove the equivalence of H^1 and H^2 for the brauer group using hilbert 90 and the SES

0->K^x->GL_n(K)->PGL_n(K)->0

And then direct limiting blah blah

[hallow bone]

sure i mean the interesting stuff requires a lot of algebra machinery or alg top

[astral ginkgo]

But the explicit construction is much nicer I think

[fierce steeple]

Uhh well here I meant H^1(X, Gm) but same result should work for GL_n, like the point is how quasicoherent sheaves work in the étale site (or other variants like fppf), namely the categories you get are all equivalent

Then for X = Spec k you have the fact that there is up to isomorphism only one vector space of given dimension

[astral ginkgo]

Well it doesn't really matter. The point is Hilbret 90 gives you an injective map H^1(G,PGL_n(K))->H^2(G,K^x)

You only get surjectivity when you take a colimit

Which you need to prove is compatible with these inclusions

[wheat meadow]

being able to explain this stuff carried my phd interview lol

[astral ginkgo]

I did my oral exam on this

[fierce steeple]

I was confused by the K^x until I realised my sins

[wheat meadow]

i always latex k^times but when not doing latex i always k*

[last talon]

PhD interviews don’t exist silly

[wheat meadow]

im still not sure whether i prefer * or vee for duals tho

[past cove]

after not being able to give an example of a genus 2 curve 😔

[wheat meadow]

indeed

[fierce steeple]

Nah i meant I am too used to expressing it as like H^1(K, Gm) or similar

[wheat meadow]

tbf i had never thought about it before

and i did in the end come up with an idea that works once you do the details

[fierce steeple]

Do a disjoint union of two genus 1 curves and say your convention has no connectedness or irreducibility conditions

[last talon]

The only example of a group that matters is BS(1, 2) (it will kill all your conjectures)

[fierce steeple]

Is this the one that kills the unit conjecture

[astral ginkgo]

H^2(G,K^x) classifies Central Simple Algebras split by K (up to Brauer Equivalence) and H^1(G,PGL_n(K)) only classifies those of degree n (dim n^2) up to isomorphism

[last talon]

Idts

[wheat meadow]

they did actually ask for irreducibility lol

[past cove]

broke: theorem 90
Woke: tate vanishing theorem

[wheat meadow]

otherwise i would have said that

[fierce steeple]

Lol

[wheat meadow]

i eventually vaguely described taking the normalization of a nodal quartic

and drew a decent picture

i guess they were happy with that

[fierce steeple]

I am confused lol why is this Tate vanishing

[last talon]

It’s a very standard (counter)example, so like if it’s reasonable to check, anything conjectured in the last however many years would’ve been tested against it

[wheat meadow]

what be BS(1,2)

[last talon]

(Funnily enough, it was a counterexample to a proof I was trying to generalise last summer)

[fierce steeple]

Bullshit(1,2)

[ornate atlas]

Naur that’s worse

[past cove]

in my mind they're the same (even tho they aren't) because they tell you when cohomology groups are trivial

[last talon]

<x, y | yxy^{-1} = x^2>

[fierce steeple]

Or do you mean like

Tate acyclicity for rigid analytic spacea

[ornate atlas]

The unit conjecture was killed by like the fundamental group of the Pete wentz manifold or something

[last talon]

(Modulo varying conventions for which way around conjugation goes, and which way around the 1 and 2 go, but all conventions give isomorphic groups)

[fierce steeple]

Tate vanishing to me is a term I feel I have mostly heard through chromatic lol

[fierce steeple]

Sure ye

I was told this group isn't too bad tbf

But yeah worse

[astral ginkgo]

The idea is the way to compare A in PGL_n and B in PGL_m is to embed them diagonally in PGL_mn. Then the induced map on cohomology will send a central simple algebra E to of degree n to M_m(E) in H^1(G,PGL_mn(K))

[ornate atlas]

Hantzsche–Wendt if I’m being less silly, but there’s more generators and relations, like 12 or 24 iirc

[past cove]

okay yeah we might be thinking of different things I don't remember anymore

[fierce steeple]

😔

[past cove]

I heard this called tate vanishing

[fierce steeple]

It's chill dawg

[past cove]

but that probably isn't the standard name

[ornate atlas]

No it’s not, you can check the counterexample by hand quite reasonably

[fierce steeple]

Probably is and I am ignorant

[fierce steeple]

Ahcokc

[ornate atlas]

I mean you shouldn’t because gap exists but you could

[astral ginkgo]

This naturally gives us the notion of brauer equivalence on the colimit. It also gives us a product descending from the kronecker product

[fierce steeple]

uSe aI

[ornate atlas]

Come again?

[fierce steeple]

[wheat meadow]

ok but gemini is like unironically decent at producing working macaulay 2 on the second or third try and this is faster than reading documentation

[fierce steeple]

Oh lol nice

Yes mc2 I find a pain lol

[wheat meadow]

assuming you know how to achieve what you're trying to achieve and can explain it properly lol

[past cove]

I've never used mc2

I've only ever used magma

💀

and it's been painful

[wheat meadow]

is painful but it's faster than sage so

[ornate atlas]

I’ve found googles AI that it automatically uses is actually great at sage for some reason and it’s like the only time I use AI

[fierce steeple]

I had an ug project where I was playing around with mc2 for weeks and then realised you could just use the koszul resolution in a clever way so I'd wasted like 7 weeks using the resolutions from macaulay2

[ornate atlas]

And I refuse to use mc2 because I’ve tried before and it was just pure pain

[past cove]

bro I'll never trust google AI it told me my flight was at terminal 7 when in reality it was at terminal 1

[wheat meadow]

oh god that's so real

[ornate atlas]

1. skill issue
2. it is generally utterly dog shit but it seems to be really good at this one thing which I shall take

At least for the very basic calculations I do in sage

[past cove]

this is what my first UG thesis project was like before I told my professor I hate this project and wanna do something else

[wheat meadow]

i will say it's good at most things in sage but it is not good at producing good plots

[ornate atlas]

If I had went with Reid it’d have been my MSc too

[fierce steeple]

Well actually I didn't realise this. The paper (whose theorem I was using to compute examples from) was updated and explained what I was trying to do basically

[fierce steeple]

Lmao

[wheat meadow]

what is reid like anyway -- i have heard many things but i have never spoken to him

[ornate atlas]

He said he and Eisenbud had noticed some patter but had no clue what it was and wanted me to work it out

[past cove]

I present to you my magma computations that took me two weeks to iron out and then my professor said "great now that you have that 10 line long isogeny find it's action on cohomology"

[fierce steeple]

Yeah Ig it is usually quite good for coding right

[past cove]

oh wait

[ornate atlas]

Brother you and and Eisenbud are the goats at random minimal free resolution and syzygy nonsense if you’re stuck what am I going to do

[past cove]

this file is an old draft

there's a longer version somewhere hm

[ornate atlas]

Yeah…

[fierce steeple]

https://tenor.com/view/suffering-gif-13429271044533447139

Hug

Sheaves.

[past cove]

tho my professor kinda saw the light fade from my eyes when he told me this and said do u wanna work on something else and I almost cried of happiness

now I'm doing even more schizo stuff

[ornate atlas]

I guess what I’ll say is that my initial experience with him was not at all positive, and from taking a course with him I think his age is possibly just telling

He’s still seemingly really quite sharp though

[wheat meadow]

makes sense

[ornate atlas]

If we come to meet in person perhaps I can say more lol

[wheat meadow]

it's kinda funny how old profs are somehow inherently polarizing, either they're really great to work with or really terrible -- i've heard very few people describe genuinely neutral experiences with old profs

anyway i should like

sleep

destroyig my sleep cycle right before my biggest block of finals is uh

a pro gamer move

but probably not good for my future

[ornate atlas]

Me too, me too

[last talon]

I can wreck my sleep schedule now, my exams are in like a month

[ornate atlas]

But also I’m beyond caring about these exams, I’m so over it and I just want to do actual maths and no longer be in Coventry

[hallow bone]

lowkey

sleep is for the weak

however, I, undoubtedly, am weak

[last talon]

Based and correct

[hallow bone]

therefore i shall sleep

[last talon]

Should I do some Riemannian geometry

[hallow bone]

you should sleep

[last talon]

I slept basically all of Sunday lol

Like, I was conscious for maybe 6 hours of that day

[hallow bone]

sunday is a track by the band sleepover disaster on their only currently released album "Hover"

[hallow bone]

one of those days huh

[last talon]

Yup

Slept through my mum calling 💀

[hallow bone]

ooff

blissfully unaware

ive had times where i forgot i had a therapy session planned next morning (even though its in my agenda) and just wake up at noon with a missed call from the therapist 💔

[last talon]

Fortunately I don’t tend to forget/miss that sort of stuff

But yeah my sleep schedule’s been fucked

[hallow bone]

oh well fragmented memory doesnt help

[hallow bone]

whose isnt lowkey 😭

[last talon]

Mine didn’t used to be

[fierce steeple]

My sleep schedule was good when I was like 11

[last talon]

Mine was good until like 4 months ago

[hallow bone]

same it all went wrong when i gained consciousness...

[hallow bone]

ah so 12 is where it all went wrong

lmaO i thought this was algechill

[last talon]

Every algebra channel if algechill if you algechill hard enough

[near lantern]

Let A be a commutative ring. Let G be a finite group acting on A. Let B = A^G. There is a functor from A-modules with a semilinear G-action to B-modules given by M ↦ M^G. Does this functor have a right adjoint?

[astral ginkgo]

A-modules with a semilinear G-action are the same as A_t[G]-modules, and then taking invariants is the same as taking Hom_A_tG, which shouldn't have a right adjoint in general

[near lantern]

Alright, so I was not thinking clearly. Let R = AG be the twisted group ring. Let triv = R/R(g-1) be the cyclic left module. It is an (AG,B)-bimodule. M^G = Hom(triv, M). So the functor has a left adjoint. It has a right adjoint iff it it cocontinuous, iff triv is projective (it is fg), iff R → R/... = triv splits. It is easy to see that a section is the same as r in R such that ∑_{g ∈ G} g⋅r = 1 (note that interestingly, this has a chance of existing even in the modular case!).

[waxen fractal]

what's A_t?

i never heard of semilinearity and now im curious

[astral ginkgo]

A_t[G] is the twisted group ring

[waxen fractal]

ah that makes vague sense

[astral ginkgo]

Same as a set but multiplication is given by (xg)(yh)= (xg.y)(gh)

[near lantern]

Try to make up the right definition of G action so both A and M and the action have a group G of symmetries.

That's really it.

[waxen fractal]

whao

[astral ginkgo]

A semilinear G-action just means g.(am)=g.a g.m

Note that any module with a semilinear G-action gives an A_t[G]-module action, and vice versa an A_t[G] module action has

(1g)(a1_G)m = (g.a g)m=g.a gm

[near lantern]

In fact, B(^op) = End_RG(triv), so if triv is projective this is an equivalence with a "block" within semilinear representations (∑_g g(r) g)R(∑_g g(r) g)-Mod. (In the case of a trivial action of G on A, this is just the subcategory of trivial A-representations. Otherwise note that triv is not actually trivial - its G-fixed points are B ⊊ A.)

[near lantern]

Let A be a G-graded ring and B = A_H its H-graded subring, where G is an abelian group and H a subgroup. What are the adjoints to M ↦ M_H? Hom_{B,H-gr}(N, M_H) = Hom_{B,G-gr}(N, M) = Hom_{A,G-gr}(A (⨯)_B N, M). (That the graded homomorphisms are the same requires separate verification.) So the left adjoint is something similar to before.

As for the right adjoint, Hom_{B,H-gr}(M_H, N) = Hom_{B,G-gr}(M, N) = Hom_{A,G-gr}(M, Hom_{B,G-gr}(A, N)). (Again the gradedness part of this requires separate verification, and in fact a graded definition Hom_{B,G-gr} if [G : H] is infinite.)

Possibly this admits a generalisation to linearly reductive group actions, but I won't think about this now. All I need currently is this right adjoint gradedCoInd: N ↦ Hom_{B,G-gr}(A, N).

The counit of the adjunction is N ↦ (Hom_B(A, N)_i = Hom_{B,G-gr}(A, N(i)) = Hom_{B,H-gr}(B, N(i)) = N_i)_{i ∈ H} = N mapping to N by the identity. So practically by definition, once again we get a reflective subcategory.

Actually the unit of the other adjunction is also an isomorphism. So it's better to say that this is a reflective and coreflective localisation.

[robust rain]

What does vakil mean by 'a long exact sequence can be factored into short exact sequences'. I don't even know what question to ask because I don't know what is trying to be said in this section. Can someone please explain what is trying to be said here

[undone idol]

if you have a long exact sequence, you can truncate terms in the sequence to give you an induced SES

because a SES is just a diagram depicting the first isomorphism theorem

so you just apply it to each successive map in the LES

which is exactly what is being done in the image

[robust rain]

hm

[wheat meadow]

Exactness of A at A^i is exactly the same as saying that
0 -> ker f^i -> A^i -> ker f^i+1 -> 0 is a short exact sequence. These are two ways of saying the same thing.

[robust rain]

I think i just need to spend more time with this. thank you for the comments though, I'll keep these in mind : )

I think the issue I'm having is that I'm not fully used to the 'categorical' definitions of kernel, cokernel, etc. Interpreting the kernel/cokernel as a morphism satisfying a univ property/as a (co)limit is kind of fucking with me

especially, when defining im f = ker(coker f))

like this feels so opaque

[lone jacinth]

@robust rain

I think the following picture is instructive

The les has been factored into (digonal) short exact sequences

[robust rain]

Quick question: what is Ki here? I assume it is being interpreted as a kernel/cokernel or whatever. But as far as I know, the kernel is not an object - it is a morphism?

[lone jacinth]

The map Ki -> Ai
is the kernel and
Ai -> Ki+1
the image

[robust rain]

Or is the kernel the pair (A, i) and this pair is univeral

[ornate atlas]

This kinda thing is also useful for say, cellular homology, it’s how we actually build that LES

[robust rain]

and are we assuming that the sequence is exact at A1 or is only a complex at A1

[lone jacinth]

You may think of kernel in just the same way you do for abelian groups or whatever.

It's a subgroup of Ai.

This is not just an object in the category, but also has an inclusion Ki -> Ai

[robust rain]

Okay. I am also coming to the conclusion that thinking of every abelian category as Mod_A may be good for me

[lone jacinth]

It being exact is equivalent to the diagonal sequences being exact

[robust rain]

oh of course that's why vakil uses the notation i for the map from the kernel - because he's trying to signify that it is an (the?) inclusion

[robust rain]

Okay so the point is that the sequence 0 -> K1 -> A1 -> im f_1 -> 0 is always exact, and if the LES is exact at A1, we are 'allowed to replace' im f_1 by K2. I think.

[lone jacinth]

Yes exactly.

You have a canonical map imf1 -> K2 and exactness is by definition that this map is an isomorphism

And then canonical isomorphism you should think of as equal, because everything is defined up to canonical iso

[robust rain]

okay thank you

I need to keep telling myself that everything is ModA

[robust rain]

can someone check my proof pls 👉 👈

[last talon]

You’re not quite done
You have to prove that \overline{h^i(x)} is actually defined (ie, that h^i(x) is a cycle)

[robust rain]

(comments on clarity/notation much appreciated)

[robust rain]

oh yes of course

This follows from x \in ker f^i, and g^i (h^i(x)) = h^{i+1}(f^i(x)) = h^{i+1}(0) = 0

[last talon]

[robust rain]

ty !!!

[lone jacinth]

I would go with $C(\operatorname{Mod}_A)$

Or at the very least
$\operatorname{Com}_{\operatorname{Mod}_A}$

[broken turtle]

jagr2808

[lone jacinth]

Not to be overly nitpicky 🤓

[robust rain]

No it's okay haha, I appreciate it. I was just following the notation in Vakil but I think I prefer C(Mod_A)

I'm becoming increasingly aware of the importance of good notation. The main obstacle for me when proving that extension and restriction of scalars are an adjoint pair was finding proper notation

[ornate atlas]

Honestly very real lol, my first category theory homework this sem I kinda bombed through just horrific notation

[rose mirage]

Do I dare ask for elaboration

[ornate atlas]

No ❤️

[rose mirage]

How could u do this to me twin 💔💔

[ornate atlas]

It was some translation groupoid stuff and tbh I just got very sloppy and ended up writing something which made sense to me at the time, and upon looking back was basically impossible to follow

IIRC I was just being a bit laissez faire about what’s a map and what’s a group element and it got hard to read. I darent share my shame

[rose mirage]

Based

[wicked gulch]

Hello. I became curious about number systems, and I was wondering if there is any visualized diagram that shows all number systems. The important part is that I want a relational chart that also includes structures beyond complex numbers, such as multicomplex numbers, split-complex numbers, quaternions, and so on. For example, I’ll upload an example image of the kind of diagram I mean. (However, I want something even more expanded.)😁

[rose mirage]

https://en.wikipedia.org/wiki/Ring_(mathematics)

c.f. https://en.wikipedia.org/wiki/Quotient_ring

[hushed bone]

This isn’t really the right channel for what it’s worth

[wicked gulch]

yes, I was even confused which channel should I go 😂

[hallow bone]

this diagram is wrong anyways

or well, not necessarily wrong

i just dislike how it denotes complements

[subtle smelt]

There are infinitely many "number systems"

[hidden shore]

What type of number are you? I choose to be a real number today.

[hallow bone]

number system horoscope

[waxen fractal]

im ω+1

(ordinal number)

[hushed bone]

That’s good, one less and you be limited

[hallow bone]

well im ω+2

[waxen fractal]

you're my successor

[hallow bone]

exactly!

[waxen fractal]

recently i had a shitpost question emerge in my mind for concrete reasons that are hard to explain without context

rather than explain those reasons, ill just ask the question

is there a natural way to encode abstract simplicial complexes as varities in UA?

[hallow bone]

you mean simplicial sets?

probably not

[waxen fractal]

nope

https://en.wikipedia.org/wiki/Abstract_simplicial_complex

[hallow bone]

oh right just a set with a distinguished down-set of P(X)

[waxen fractal]

yea

[hallow bone]

yeah no probably not

similar to clones i dont see why you couldnt realize them as heterogeneous algebras though

you can realize simplicial sets this way

and an abstract simplicial complex is nothing more than a simplicial set where each n-graded element is determined by its n-1-graded components

[waxen fractal]

yea multisorted stuff is kinda based

if that's what you mean

[hallow bone]

heterogeneous UA works essentially the same as regular UA

[waxen fractal]

that's what i heard

[hallow bone]

i do yes

[waxen fractal]

something I've observed is that if you define quasigroups heterogeneously, you recover the notion of isotopy

i.e. maps
A × B -> C
B × C -> A
C × A -> B
that induce isomorphisms A ≅ B ≅ C

[hallow bone]

ive never heard of that

interesting

[waxen fractal]

I've made sure to go through the details to make sure they're equivalent notions

[waxen fractal]

yea i just discovered it myself so i have no reference

but surely some quasigroup theorists must have realized this, i wouldn't know cuz i never studied them

[hallow bone]

i suppose this has to do with the fact that a latin square can be seen as a shape that has the shadow of a cube from all three axial directions

[waxen fractal]

ah yea cuz they're just presheaves

no need to even pull out the lawvere theory

wait ok well here's the distinction

im thinking of encoding each ASC as its own variety

for nefarious shitposting purposes

ok ill give the context

it's literally 100% for shitposting reasons

[hallow bone]

asc?

abstract simplical complex

[waxen fractal]

like there is no serious math here

it's just "what if we could triangulate manifolds and study them with UA to troll this guy"

[hallow bone]

did you know affine varieties do correspond to varieties in the UA sense

[waxen fractal]

no

wha

[hallow bone]

well

a commutative ring can be faithfully identified with its theory of modules

and morphisms of varieties will correspond to ring morphisms because by a uniqueness result of McKenzie the group structure of modules will always be sent to the corresponding group structure in the other variety of modules

[waxen fractal]

wait hold on

that's kinda cracked

rings are varieties

and affine varieties are rings

[hallow bone]

lol

yes

[hushed bone]

This is actually really obvious cuz both of them are called varieties and are algebraic

[hallow bone]

this guy gets it

[waxen fractal]

chair monkey w the elite ball knowledge

[hushed bone]

Mathematicians don’t use the same name on things unless they r the same

[hallow bone]

yes

[waxen fractal]

yea, mathematicians are famously normal people

they're also very regular and nice

ok can we complete the following pushout square of concepts:

commutative theories
^^^
rings >>> schemes

these are the two perspectives on rings i commonly have

[hushed bone]

Yeah, but are they quasicompact quasiseparated ind-étale?

😂😂😂😂

[hallow bone]

i know a paper that develops algebraic geometry and scheme theory for commutative theories

[hushed bone]

You can’t make a pushout square cuz the arrow goes schemes -> rings cuz it’s contravariant

[waxen fractal]

wait no that's not how it works

rings >>> scheme^op

[hallow bone]

https://arxiv.org/abs/0704.2030

[hushed bone]

Bro said he makes a theory to generalize rings and schemes we already GOT THAT NOW

You made the wrong theory lil bro

[hallow bone]

damn this was uploaded almost exactly two weeks before my birth

[hushed bone]

What

Say sike

[hallow bone]

i never jest

[waxen fractal]

sometimes i forget teens exist

[hallow bone]

💔

am I that forgettable

[waxen fractal]

nah just the concept of teens

anyway

[waxen fractal]

this is going straight to my bookmark bar

[lyric rapids]

I guess im forgetable.

[waxen fractal]

you're the concept of teenage??

[lyric rapids]

Yes i am the one that created the concept of teenager.

[waxen fractal]

john teen

[distant harness]

N = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 20, 21, 22, 23, ... }

[lyric rapids]

Everyone had always been in their 20s before i create them.

[hallow bone]

geometry dash was based on the horrifying story of Joe Metri Desh

[hallow bone]

[limpid horizon]

Lol

[hidden shore]

LOL

[ornate kindle]

So I know for a simple Lie algebra its finite dim reps are completely reducible, and its irreducible reps are characterized by their highest weight vectors

It follows then, that if $V_\lambda$ is a irrep with highest weight $\lambda$, we have
$V_\lambda \otimes V_\delta \cong \bigoplus_i V_{\gamma_i}$

[broken turtle]

NotABot

[ornate kindle]

Can anyone point me towards resources that could help me derive the set of γ_i weights on the right, given the weights λ and δ on the left?

And given a specific type of simple Lie algebra, of course

I'm currently looking at Victor Kac's "Infinite Dimensional Lie Algebras" and Fulton and Harris's Representation Theory, but if there's other resources that are better for that I'd love to hear it

[hallow bone]

I don't think you're gonna get far with looking atn infinite dimensional Lie algebras lol

since I assume you want your lie algebra to be finite dimensional?

that's where all this nice stuff happens at least

[ornate kindle]

It's more a book I already had on hand, but it's got a lot on characterizing integral representations of Kac-Moody algebras via weights, of which the finite-dim case is a subset

And yeah, I'm specifically working with simple Lie algebras at the moment

[hallow bone]

I can't seem to find much which leads me to believe it's a hard problem in general (which I have no problem believing, to be honest)

[ornate kindle]

I know, for example, that the highest highest weight is gonna be λ+δ, and that all the weights in the space are gonna be the sum of a weight from V_λ plus a weight from V_δ

[hallow bone]

but maybe someone like jagr will have a proper answer for you

[hallow bone]

yeah that's simply from the weight decompositions right?

[ornate kindle]

That plus the usual coproduct of a Lie algebra yea

[hallow bone]

wdym?

[ornate kindle]

Uh, if v is a h.w vec in V_λ and w a h.w vec in V_δ, then

[broken turtle]

NotABot

[hallow bone]

right, yes
coproduct in that sense

[ornate kindle]

Yea

[hallow bone]

gotta love Hopf algebras

[ornate kindle]

Mmhm!

[hallow bone]

but yeah this

[ornate kindle]

Well for now ill keep reading and see if anything jumps out at me, or if I see a different angle of attack for my problem

[hallow bone]

apparently fulton and harris says a lot about this

[ornate kindle]

Yeah I saw some stuff that looks pretty relevant in there, but it's currently a bit over my head. I'm working on the preceeding chapters right now

Good to know that I'm not working up to something completely unrelated :P

[wary elbow]

You're looking for branching rules

In general, yes, this is a complicated problem, but my understanding is that the Littelmann path model / standard monomial theory gives a full solution, at least theoretically

[ornate kindle]

Ah cool, thanks!

A quick google search is giving me a lot of stuff I'm not familiar with, particularly lots of physics stuff, but that does look pretty relevant :)

[wary elbow]

Yeah, it's an involved combinatorial framework, but it does provide a full solution. See eg Corollary 1 in this paper:
https://www.mi.uni-koeln.de/~littelma/papers/cambridge.pdf

[ornate kindle]

Whoa

Yeah I'm having trouble parsing most of that, but what I do understand at a glance makes me think that paper ties to a bunch of different aspects of my research! Thank you so much for sharing!

[wary elbow]

happy to help!

[hallow bone]

that is very sleek

might actually start liking combinatorics because of lie theory

[lost ingot]

;-;

[wary elbow]

yeah, I was mad at first because I felt like I got suckered into combinatorics when I chose to study representation theory, but I've made peace with it

[hallow bone]

take the first lesson, don't pay afterwards

[hallow bone]

I can make peace with it if there are pretty pictures

(which there usually are)

I usually do universal algebra, and recently started diving into pure commutator theory because I needed some results from there and suddenly had to formalize very combinatorial stuff with like hybercubic sets and n-dimensional relations and whatnot

but the pretty pictures made it worth it

[ornate atlas]

No, but please dm @alpine dirge about this please (and in any similar situations in the future!)

[lost ingot]

k

[hidden shore]

[sly rune]

<@&268886789983436800>

[robust rain]

Sanity check, the restriction maps are as follows:

Let $U \subset V$. Then $\text{res}{V,U} : i{p, *}(V) \to i_{p, *}(U)$ is given by sending ${e} \mapsto {e}$, and $S \mapsto S$ if $p \in U$, and $S \mapsto {e}$ otherwise.

[broken turtle]

swifteeee

[robust rain]

I feel like I'm misunderstanding something, but I cant quite tell what

[lone jacinth]

It's just the identity map on S, yes

[sacred sentinel]

If by $S \mapsto S$ you mean the identity, then yes

[broken turtle]

Moorts

[robust rain]

right, because a presheaf is a contravariant functor Op(X) \to Set, so restriction maps are morphisms in Set.

So it's either the identity id_S or the unique map S \to {e}

[lone jacinth]

(or the identity on {e})

[sacred sentinel]

I mean technically there could be other choices, but these are the canonical ones

[robust rain]

okay thank you

Can I ask too, what is the point of the skyscraper sheaf?

why does vakil label it as important

[sacred sentinel]

It usually comes up in exact sequences induced by the inclusion of a point, e.g. in the proof of the riemann roch theorem

Like for any map of schemes, you can pull back sheaves along that map and in case of the inclusion of a point, the pullback will be a skyscraper sheaf

[robust rain]

Hm. At this point these words mean nothing to me, so I will take it on faith that this is important haha

thank you

[sacred sentinel]

Ehm wait that's not quite right, im mixing up directions, the pushforward of a sheaf along that map will be a skyscraper

[lone jacinth]

One nice property is that
Hom(F, Sky(S)) = Hom(stalk of F at p, S).

This you can use to show that categories of sheaves of modules have enough injectives for example.

[robust rain]

I'll keep this in mind for when I better understand all this haha. There is so much information contained in every word, it feels a little overwhelming

[timber patrol]

ag in a nutshell:

[lone jacinth]

I guess intuitively the fact that the skyscraper sheaf only cares about the point p simplifies things to where you only need to worry about the point p

[robust rain]

Gotcha. Thank you

[void plank]

I have a question about exercise 5.1.1 in Weibel

I got as far as computing H_(p+q) in this stack exchange answer, but I'm confused about the last equality on that line. https://math.stackexchange.com/questions/983722/total-complex-homology-exact-sequence

E^1_(p-1,q+1) is the vertical homology at (p-1, q+1). But the equality written in the numerator doesn't imply a is in the kernel of d^v_(p-1,q+1)?

[hushed bone]

Maybe this is earlier than what I’m gonna say, but does this not follow from the fact that the spectral sequence collapses on E^2 and then looking at the infinity page?

And I mean the infinity page is just E^2, you use that obviously

Anyway uh, this is way too in the weeds for me to untangle

But is there a chance that a is in the kernel just by definition?

[void plank]

Yeah, I think the spirit of the problem is just to manually write out each of these three terms and compare

[hushed bone]

If a is an element in an E^2, the stuff there is homology from stuff on an E^1

So by definition it’s in the kernel of some differential

I don’t want to really go through and see which differential you’re referring to cuz my eyes glaze over with all this crap

[void plank]

The exercise seems to want (a,b) in E^0_(p-1, q+1) x E^0_(p,q), hence my confusion

[hushed bone]

Idk

Sorry like I said this is way too in the weeds for me

I am not convinced it’s worth doing this sort of nitty gritty stuff for homological algebra yourself haha

[void plank]

But I like spectral sequences 🙁

(that collapse on E^2)

[hushed bone]

Yeah I do too, so I just use the E^infty page

[lone jacinth]

An element of E^0_p,q that maps to 0 vertically you can think of as an element of E^1. And then E^1_p,q maps onto E^2_p,q giving you the desired map

[void plank]

Right, but the user on the stackexchange post has this for their expression for H_(p+q). Based on the numerator, a doesn't necessarily have to be in the kernel of the vertical map right?

I guess the point is that once you get to the first page, you only care about the horizontal maps? Seems that way

[lone jacinth]

So the image is like this I guess

So if b is 0, then d^v(a) = 0 and we can think of a in E^1

And then you get
E1_p-1,q+1 x ker d_h / (im d_h, 0)

Where you think of d_h is a map in E1

[void plank]

To clarify, d^h(b) = -d^v(a) implies b is in ker(d_h: E^1 --> E^1) since ker(d_h: E^1 --> E^1) consists of elements that are in the image of d^v at (p-1, q+1) right?

[lone jacinth]

Since E1 is the homology of d^v, being in the image of d^v is being 0

[void plank]

Wait, so am I wrong?

[lone jacinth]

I mean, kinda. Or your wording is weird.

The kernel are those elements sendt to 0

Which is different from those elements that are 0

[lone jacinth]

Like d^v at p-1,q+1 takes things to p-1,q but the kernel lives in p,q

[void plank]

So I agree d^v_(p,q) (b) = 0 implies b is in E^1_(p,q). The kernel of d^h_(p,q): E^1_(p,q) -> E^1_(p-1,q) are those equivalence classes in E^1_(p,q) sent to [0] in E^1_(p-1,q).

I'm assuming d_h acts as d_h([x]) = [d_h(x)]?

Ah ok, in which case d_h[b] = [d_h(b)] = [-d^v(a)] = [0], so [b] is in ker(d^h: E^1 --> E^1)

[lone jacinth]

Yeah, so the point is that b always defines an element of E^2_p,q.

This then gives a map H_n -> E^2_p,q.

Then the question is what the kernel of the map is. That would be when [b] is 0 in E^2, which happens when [b] is 0 in E^1, which means b = d^v(y). But then we can just replace (a, b) with (a', 0) in H_n.

In which case d^v(a') = 0, so a' is in E1 and the remaining relations is are the same as for E^2

[void plank]

I see, my mistake was trying to see how each relation gave an individual piece of the isomorphism

Thanks!!

I also had a quick question about the next exercise in Weibel. From the definition of E^2_(p,q) directly, its easy to see the first two conditions. What I'm having trouble proving are the relations he gives

For example, the definition of E^2 has a quotient by im d^h_(p+1,q). So if you take c such that d^v(c) = 0, its clear that d^h(c) will be in im d^h_(p+1,q). Why does this imply a = 0 though? Surely any a in ker d^v will also work?

To clarify, it's easy to see that these pairs are in ker d^v. I'm confused why these are the only ones

[lone jacinth]

Why are you wanting a=0?

I'm not understanding what you're asking

If b = d^h(c), then
d^h(b) = 0, hence d^v(a) = 0 so (a, 0) is a relation

[void plank]

Ah whoops, I was trying to show that I could specifically get (a=0, b=d^h(c)) by examining im d^h. But I guess we can add these relations together so it doesn't matter

[void plank]

Is the second part of the problem just an application of the previous exercise?

[lone jacinth]

It's similar at least, if not a direct application

[void plank]

I guess my only concern is that now we're not assuming there are only 2 non-zero columns. The total differential will be d_v + d_h on each piece of the direct sum in a particular degree of the total complex.

But I suppose the point is still that I'm determining a graded piece of the homology of the total complex up to extension?

[lone jacinth]

Yeah, that's why you need the extra condition of d^h(a) = 0

[void plank]

And checking the map in (3) is well-defined just amounts to checking that its independent of the relations in step 1 right?

[silver goblet]

i am looking at this problem in Atiyah-Macdonald (ex 5.2). I know how to solve this using the Going Up theorem, but I am curious about whether this can be done with Zorn's lemma. You can consider the poset of subrings R between A and B with lifts of f, where R_1 and a lift f_1 is \le R_2 and a lift of f_2 if the obvious diagram commutes. its not empty, and for a chain you can just take unions of subrings to bound it, so the poset has a maximal element. What I want to do is say: if x is not in the maximal subring R in the poset, then it satisfies some polynomial, and you can send x to a root of that polynomial in Omega, which exists via algebraic closure. However, this may not work: x may not be algebraically independent from the rest of R -- is there some way to modify this argument to make this work?

[hallow bone]

you'd essentially be proving part of the going up theorem no?

[void plank]

Why is this the case? Surely we need the rows of the complexes to be exact to apply the 5-lemma. That's not necessarily the case with these spectral sequences, right? I must be missing something

[hushed bone]

Lol, there was discussion about this a week or two ago

#advanced-algebra message

It was back here, but I think ultimately you end up using going up

The main problem is reducing to the case of integral domains

[lone jacinth]

This result immediately implies going up, so proving it without going up can't be easier than just proving going up

[astral ginkgo]

The problem with using Zorn's ended up being non-uniqueness of minimal polynomials for integral extensions (even in the case of domains)

[lone jacinth]

im d -> ker d -> next page -> 0 -> 0
is exact
So just use that I guess

[void plank]

Ahh, my bad. Thanks!!!

[vague pawn]

Let P,Q be prime ideals in a commutative ring A,
if A_P \cong A_Q (localizations) is it true that P=Q?

Is the proof just: every element of P is invertible in A_Q, so its either invertible in A or in Q, but P cannot contain units so P \subseteq Q.

Is this right?

[fierce steeple]

No, e.g. take A = Z x Z and consider the prime ideals generated by e_1 = (1,0) and e_2 = (0,1)

[near lantern]

Surely it means as A-algebras?

Well not surely, but maybe.

[fierce steeple]

Sure I mean this should be quite sensitive to how it is stated

[vague pawn]

I have an isomorphism of rings...

[near lantern]

And this isomorphism doesn't respect the map from A?

[vague pawn]

I need to check if it does xD

if it does what does it change?

[vague pawn]

also what is wrong in this proof

[near lantern]

If it doesn't, it's false if there is any automorphism of A that maps P onto Q.

[fierce steeple]

I don't really get what you are doing

[near lantern]

To look at elements of P as elements of A_Q, you use the map from A to that.

[vague pawn]

oh right

[near lantern]

Also typos (any element of A\P is invertible in A_Q, so it lies in A\Q, and vice versa), but these are less important.

[vague pawn]

showing two prime ideals are equal by showing the stalks of SpecA at the two points are isomorphic

[fierce steeple]

But also this is sort of a "wrong" question in that you are going from isomorphism to equality

[near lantern]

It's really not that wrong.

[fierce steeple]

Well sorry I mean more like why I wouldn't expect it to be true

[near lantern]

Equality is really just isomorphism preserving enough structure.

[near lantern]

It is true if you assume they're isomorphic as A-algebras.

So really just a question of whether you tracking enough structure.

[fierce steeple]

Ye sure

I guess point is that you can remember p as the preimage of the maximal ideal of pA_p?

[near lantern]

Yes.

[vague pawn]

Another question:

Is there a way to recover how a morphism of affine schemes acts on points from how it acts on global sections?

I am trying to do it via stalks. is there another way?

[fierce steeple]

Yes, the global sections determine everything for affine schemes. The standard way to prove this is to work carefully with stalks

[vague pawn]

great

[fierce steeple]

This is smth covered in any AG book but I imagine you are doing it as an exercise or smth aha

[vague pawn]

of course xD

[lone jacinth]

The points are prime ideals of the global sections. So that's how they're recovered

[vague pawn]

this is one way to recover this

but I want to prove uniqueness

[lone jacinth]

Uniqueness of what?

[vague pawn]

of the original morphism

There shouldn't be two morphisms that act the same on global sections

[fierce steeple]

If you can recover the map from the global sections then it's uniquely determined by the global sections

[lone jacinth]

So what you meant to ask is if it's possible to recover how it acts on global sections from how it maps points

[fierce steeple]

Well okay you will only recover how it acts on the space by what was just said so u also need the map of sheaves ofc

[vague pawn]

thats easy

its part of the definition of a morphism of ringed spaces

[fierce steeple]

Ngl I am finding this hard to follow when you say "original morphism" since there are two directions lol

[lone jacinth]

Then I have no idea what you're asking

[fierce steeple]

But if you know how it acts on global sections then by what jagr said you can recover the map on points

So now you just need to worry about the map of sheaves

[vague pawn]

I have a morphism of ringed spaces $\phi:SpecB \rightarrow SpecA$. I know it induces a map of rings $\pi: A \rightarrow B$, and this induces a morphism of ringed spaces $\psi:SpecB \rightarrow SpecA$. How do I know that $\phi=\psi$?

[broken turtle]

ExpertEsquieESQUIE

[vague pawn]

This is what I mean

[lone jacinth]

I see, so you want to show that phi(p) = pi^-1(p) just from the definition of morphism of locally ringed space?

[vague pawn]

yes

I thought of using the stalk at P of B, which should be isomorphic to A_phi(P) and A_pi^-1(p)

[lone jacinth]

Well it shouldn't be isomorphic, but you have the commutative diagram
A ---------> B
v v
A_phi(P) -> B_P

So for the bottom to be a map of local rings you need phi(P) to be a subset of pi^-1(P) at least.

And a ring map sends units to units, so anything not in phi(P) isn't mapped to P by pi. So phi(P) = pi^-1(P) boom boom

I think you need to consider morphisms of locally ringed spaces though.

[fierce steeple]

Ye

[vague pawn]

the exercise uses only ringed spaces

and not lrs

[lone jacinth]

What does the exercise say then?

[lone jacinth]

Hmm, okay so you have to use that both specA and specB are affine.

[vague pawn]

so here in the bottom row we have an isomorphism

and also another diagram like this with A_phi(P) replaced by A_pi^-1(P)

[lone jacinth]

You don't get an isomorphism there. Just consider when A and B are fields for example

[vague pawn]

I don't understand

[lone jacinth]

If A and B are fields then B_P = B and A_phi(P) = P.

But not every homomorphism between fields is an isomorphism

[vague pawn]

this one is just the homomorphism induced from SpecB --> SpecA, and since this morphism of ringed spaces is an isomorphism, this homomorphism should also be an isomorphism,

[lone jacinth]

Oh, I didn't catch that specA and specB where isomorphic

Then it's not hard

Isomorphisms take non-units to non-units

I thought you were trying to give the correspondence between morphisms specB -> specA and ring maps A -> B

Reading is hard

[vague pawn]

I was trying to at the beginning

how much more work is it?

[digital parcel]

affines Spec A are always locally ringed spaces (the stalk of O_{Spec A} at a point P in Spec A is just A_P, which is local). if you are given a priori that Spec A -> Spec A' is an isomorphism, then the map on stalks is a local ring homomorphism (by what jagr said), so you have a morphism of locally ringed spaces in this case

[vague pawn]

then the map on stalks is a local ring homomorphism
you meant iso here?

[lone jacinth]

Not sure.

The hard part would be to establish that a map of ringed spaces specB -> specA is a map of locally ringed spaces

[digital parcel]

yeah it's also an isomorphism, my point was just that it is a morphism of locally ringed spaces

[lone jacinth]

No, just a map of local rings (a ring homomorphism that takes non-units to non-units)

But iso would do

[vague pawn]

so in the iso case the bottom map here is an isomorphism and local

right?

otherwise we need to do some work to show its local?

[lone jacinth]

Yeah

[vague pawn]

ok

enough for now

[digital parcel]

somehow I don't feel like a general ringed space morphism Spec B -> Spec A should be a morphism of locally ringed spaces

but i can't think of an example

[lone jacinth]

Oh yeah,
k[[x]] -> k((x))

[digital parcel]

Oh yeah

[lone jacinth]

Then you can map the point of spec k((x)) to either point

[vague pawn]

ok this actually makes a lot of sense

we an get rid of the local condition in the iso case because iso -> local

but if we don't have iso we need the morphism to be a morphism of LRS

[digital parcel]

I did remember thinking it was a little weird to define isomorphisms before general morphisms when i first read that chapter lol

but i guess it's fine since he only cares about defining what an affine scheme is in that section

[vague pawn]

thanks @fierce steeple @lone jacinth @digital parcel

cya

[lone jacinth]

I guess that makes it more of a "misleading exercise" than an "enlightening exercise"

[digital parcel]

maybe he has some warning saying "morphisms aren't always ones of ringed spaces" or something can't remember

i guess he does say (which can be strangely confusing)

so it's both misleading and enlightening

[vague pawn]

not at this point

[digital parcel]

hmm

oh whale

[elfin ice]

It's me again back here, this time trying a more advanced course this semester:
I face a final project, with the topic of: Satake isomorphisms and geometric Satake
with relevant references given:

Cartier, Representations of p-adic groups
Gross, On the Satake isomorphism
Mirkovic and Vilonen, Geometric Langlands duality and representations of algebraic groups over commutative rings
I read through and understood most content of the first two papers, but for the last one I feel completely confused. Any help, such as recommending papers to read in a given order, are welcome, thank you!

[eager hound]

Yeah the last paper is a big step up from the others.. I guess to understand it it would be useful to have a basic understanding of perverse sheaves and affine grassmanians, I think these notes https://arxiv.org/pdf/1603.05593 are a very nice introduction to the geometry of the affine grassmanian

And section 5 gives a pretty minimal presentation of the proof of satake

[elfin ice]

ah, a classmate also recommended this for me
will add this into the list of important references

[void plank]

I'm having some trouble parsing this proof in Weibel's hom alg book.

As an example, I have in mind C = Total complex of some double complex with the usual ascending vertical filtration. The d in the proof would then be the differential of the total complex if I'm not mistaken

He defines A^r_p to be those c in F_pC such that d(c) is in F_(p-r)(C) and then considers its image under the surjection F_pC -> F_p(C)/F_(p-1)(C) = E^0_(p,*)

But if we're considering the total complex as an example, the elements that map to F_p-r under d_tot are those in columns p-r+1 and below. So wouldn't their image under the above surjection be 0?

[astral ginkgo]

First question: Did you check the errata

[void plank]

No errata for this page in the book 🙂

[subtle smelt]

All spectral sequences originate from the dead sea scrolls, left over from ancient higher beings. The human mind was simply not made to understand them this explicitly.

[idle copper]

I have a question , if gauss elimination of a given inner product is in fact an orthogonalization of a given base
like the fact that we get a diagonal matrix means the new base is orthogonal, for some reason the AI was being contrarian and insisting it is wrong

so i am doubting if my intuition is right

[warm owl]

So I get that the representation theory of SU(2) is essentially accomplished by picking out a U(1) subgroup and examining the action of SU(2) generators on the restriction to said subgroup--is this how the procedure works for arbitrary SU(n) as well or?

[lone jacinth]

I'm not sure I understand your question.

You have an inner product given by x^T A x for a matrix A (?)

Then you want to do gauss elimination on A (?)

This will just result in the identity matrix. Not sure which basis you're asking about.

[elfin ice]

if you want to get an orthogonal basis of an inner product, the Gram-Schmidt process and matrix congruence is what you might need:
Gauss elimination does no help to that.

[idle copper]

ah yeah i know about gram schimidt, but i want to point that it is possible using gauss elimination

[distant harness]

I don't think it is.

[idle copper]

that is what it mean to have an orthnormal basis ( iguess ) relatively to the inner product

because the matrix of the product is given by the value of basis vectors with the said product

[lone jacinth]

But row reducing A is not changing the basis for the inner product

[distant harness]

I suppose you can say that Gaussian elimination does produce a matrix that represents the inner product over an appropriate basis -- for the boring and irrelevant reason that Gaussian elimination always produces the identity matrix when you start with a full-rank matrix!
But it doesn't really count as doing what you claim, because Gaussian elimination will tell you nothing about what the appropriate basis is.

[idle copper]

it does through the independent functionals

[distant harness]

I'm not sure what you mean by that, but I'll stick out my neck and say you're wrong.

[idle copper]

i need to think about this

[idle copper]

i ll try to work the example that did lead to all this questioning in my head probably so we can share the same perspective

so we get only the square part and all those linear forms given by the elimination are independent

[distant harness]

Huh, I don't see any Gaussian elimination going on there.

[idle copper]

i honestly just translated the name

the literal translation
is gauss reduction for quadratic functions

where you write a quadratic function as the same of the squares of linear forms

i just want to mention that in the second image the quadratic forme is the some of squares only if the vector basis are orthogonal

relatively to the associated bilinear form

[distant harness]

Yeah, that's not what "Gaussian elimination" means in English.

it looks like there isn't even an article about Gaussian elimination in French Wikipedia(!).

(Which is weird because "row echelon form" sure sounds like it's coming from French).

[idle copper]

yeah true

[distant harness]

Hmm, here it is, it just doesn't have interwiki links: https://fr.wikipedia.org/wiki/Élimination_de_Gauss-Jordan

[idle copper]

because what comes to my mind with gauss elimination is usually the gauss method to inverse matrices
or to make the bellow diagonal all zeroes

[lone jacinth]

Anyway, for a change of basis matrix P, the quadratic form in the new basis will be
P^T A P

If this is the identity, then P is an orthonormal basis. This method appears to only produce an orthogonal basis, but you can just rescale that assuming A was in fact positive definite to begin with.

[idle copper]

yes exactly

this is what i want to confirm before mentioning this to my students

that they can do gram shmit algorithm
or even use the gauss elemination to get orthogonal base and just divide by the norm to make it orthonormal

and honestly quadratic reduction by gauss's method seems to be way easier calculation wise than gram schmit

[distant harness]

I suppose if we start with a positive-definite real symmetric matrix we can do something like Gaussian elimination, but each time we do a row operation we also do the same column operation, clearing out the top row at the same time as the left column, and proceeding by induction. When we need to make a pivot element 1 we can do it by adding in half of the difference from another nonzero element in the row/column, dividing both its row and its column by the square root of the element (since the diagonal elements of a positive definite matrix are always positive).

[idle copper]

true

this is actually exactly the case i am thinking about sinc i am starting with an inner product which is positive and symetric

[distant harness]

I suspect that if done systematically, this ends up with the same result as Gram-Schmidt applied to the standard basis, just with the calculations arranged differently. Perhaps less work too since the matrices get smaller during the recursive process?

[idle copper]

true , because gram schmidt the calculation gets heavier as the dimension increase the number of inner product one need to calculate increases drastically

[distant harness]

On the other hand, if what we start with is an inner product that's not given as a matrix, but through some other process (such as an integral for a function space), then we'd need to compute the same number of inner products in order to matrixify it before we can get off the ground.

[idle copper]

true

yeah exactly i forgot about that part 😆

[elfin ice]

sounds like some numerical linear algebra stuff, hmm

[hallow bone]

Let L be a finite dim lie algebra over an alg closed field (more generally any vector space endowed with a bilinear multiplication). Then, let Λ be the additive submonoid generated by the eigenvalues of some derivation δ, and ⊕_λ∈Λ L_λ the primary decomposition

It happens that [L_λ, L_μ] ⊆ L_λ+μ, so L is naturally graded by Λ (even though almost all of the components will be 0)

is this useful at all? common knowledge? its something interesting at least

[drowsy niche]

yes and yes

this is used repeatedly in humphreys iirc

ok hm not quite